Math 105/206 - Quiz 1
16 Jan 2015
Write your name AND student # somewhere on this sheet.
No calculators or books. Please write down the steps of the solution (not
just the result) to get full marks. (10 marks total)
Problem 1
(a) Write down a normal vector for the plane with equation 2x−5y+z = 1−x−y
(2 marks).
Solution
The equation can be rewritten as 3x − 4y + z = 1, so one normal vector is
given by h3, −4, 1i, since a normal vector to ax + by + cz = d is ha, b, ci.
(b) Find an equation of the plane parallel to the plane 2x − 5y + z = 1 − x − y
and passing through the point (0, 0, 0) (2 marks).
Solution
Since parallel planes have parallel normal vectors, this amounts to finding
the equation of the plane with normal vector h3, −4, 1i passing through the
origin O = (0, 0, 0).
−−→
For P = (x, y, z) we have OP = hx, y, zi, so the equation is
h3, −4, 1i · hx, y, zi = 0
and after computing the dot product
3x − 4y + z = 0.
Problem 2
Given the function f (x, y) = x3 + xy + y 2 :
(a) Calculate the partial derivatives fx and fy and find the critical points (2
marks).
Solution
The derivatives are
fx = 3x2 + y
and
fy = x + 2y.
Since the derivatives exist everywhere, the critical points are just the ones
such that fx = fy = 0. From fy = 0 we get x = −2y, and fx = 0 becomes
1
. In the first case x = 0, and in
12y 2 + y = 0, with solutions y = 0 and y = − 12
1
1
the second x = 6 . Thus the critical points are (0, 0) and ( 61 , − 12
).
(b) Calculate the second-order partial derivatives fxx , fyy , fxy , fyx (2 marks).
Solution
We have
fxx = 6x
fyy = 2
fyx = 1
and fxy = fyx by Clairaut’s theorem (or by direct calculation).
(c) Use the second derivative test to classify the critical points (2 marks).
Solution
Let us calculate the discriminant
2
D(f ) = fxx fyy − fxy
= 6x · 2 − 12
= 12x − 1.
1
Now D(f )(0, 0) = −1 < 0, so (0, 0) is a saddle point, and D(f )( 61 , − 12
) = 1 > 0,
1
1
so ( 6 , − 12 ) is a local extremum.
1
1
Since fxx ( 16 , − 12
) = 1 > 0, the point ( 16 , − 12
) is a local minimum for f .